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An Old Quantum Mechanical Analysis of
a Simple Model of the Alpha Particle

An alpha particle, composed of two neutrons and two protons, is an amazing structure. It is relatively compact and has an extraordinary level of binding energy compared with smaller nuclides such as a deuteron or triteron. Binding energy is like, and perhaps is identical with, potential energy. In a previous study of deuteron-like nuclei it was found that the kinetic energy is proportional to the fourth power of the geometric mean of the cluster size. Thus the alpha particle with clusters twice the size of those of a deuteron should have kinetic energy 24=16 times that of a deuteron. On the other hand a H3 or He3 nucleus with mean cluster size √2 times that of a deuteron should have kinetic energy 24/2=4 times that of a deuteron. The relative sizes of the potential energies and binding energies should also be roughly 16 and 4 times that of a deuteron. The binding energies due to the formation of nucleonic pairs also have be taken into account but the binding energies of the deuteron of approximately 2 million electron volts (MeV), that of H3 and He3 at about 8 MeV and that of the alpha particle at 28 MeV at in roughly those proportions.

What is carried out here is an analysis of a simple model of an alpha particle along the lines of Neils Bohr's original analysis of a hydrogen atom.

A visualization of the first model is given below.

For simplicity in the following it is assumed that neutrons and protons have the same mass, m.

The two neutrons form a spin pair, but since they repel each other, they a moved to the limit of the distance at which spin pairs form. In effect there is a potential well due to the spin pair formation, as illustrated below. The separation distance for the centers of the neutrons is denoted as d. The same applies for the pair of protons.

If the potential well of the pair bond is combined with a potential energy that is inversely proportional to the separation distance the following picture results.

The above indicates why the separation distance between the neutrons and between the protons is determined characteristics of spin pair formation and is left out of the analysis.

Let r be the distance of the centers of the nucleons from the axis of rotation of the system. Let v be the tangential velocity of the nucleons. The angular rate of rotation is denoted as ω.

Now s will be used to denote the perpendicular distance between the neutrons and protons. Since r=s/2 and v=ωr=ωs/2 the quantization of angular momentum

4m(ωs/2)(s/2) = mωs² = nh
and thus
ω = nh/(ms²)
ω² = n²h²/(m²s4)

Dynamic Balance

The so-called centrifugal force on a nucleon is mv²/r which reduces to mω²r, or equivalently, mω²s/2. Dynamic balance requires the attractive force on a nucleon balance the centrifugal force. For this initial analysis only the attractive force due to the closest nucleon of the opposite type will be considered. Thus,

mω²s/2 = Hq*exp(−s/s0)/s²
and hence
ω² = 2Hq*exp(−s/s0)/(ms³)

Equating the two expressions for ω²

2Hq*exp(−s/s0)/(ms³) = n²h²/(m²s4)
which reduces to
s*exp(−s/s0) = n²h²/(2mHq)
which can be expressed as
(s/s0)*exp(−s/s0) = n²h²/(2mHqs0)
or, equivalently as
σ*exp(−σ) = n²h²/(2mHqs0)

where σ = s/s0.

The expression on the left is a nonlinear function of σ that rises from zero for σ=0 to a maximum of exp(−1) for σ=1 and then descends asymptotically toward zero as σ increases without bound.

The upward branch from σ=0 can be approximated by γσ where γ=exp(−1). Thus the relative separation distance σ and separation distance s are quantized as

σn = n²h²/(2mHqs0γ)
sn = n²h²/(2mHqγ)

Since ω = nh/(ms²) this means that ω is quantized as

ωn = nh/(ms²)
ωn = (2mHqs0γ)²/[(nh)³m]
which reduces to
ωn = 4m(Hqγ)²/(nh

Tangential velocity is given by 2nh/(ms) so it is quantized as

vn = 2(2mHqγ)/(mnh) = 4Hqγ/(nh)

The kinetic energy K of the system is 4(½mv²) so

Kn = 2m[16(Hqγ)²]/(nh)² = 32m(Hqγ/h)²/n²

Thus the kinetic energy of an alpha particle is inversely proportional to the square of its quantum number n.

The Potential Energy of the System

The potential energy of a neutron separated a distance s from a proton is

V = ∫sFdz
or, with the formula presumed
V = Hq∫s[exp(−z/s0)/z²]dz

A change of the variable of integration from z to ζ=z/s0 puts the potential energy function into a more suitable form; i.e.,

V = (Hq/s0)∫σ[exp(−ζ)/ζ²]dζ

where σ=s/s0.

The term ∫σ[exp(−ζ)/ζ²]dζ can be approximated over some range as Γ/σν where ν>1. Thus potential energy is quantized as

Vn = HqΓ/snν
which evaluates to
Vn = HqΓ(2mHqs0γ)ν/(nh)

The Net Radial Force on a Nucleon Due to
Both of the Nucleons of the Opposite Type

The distance to the nearest nucleon of the opposite type is s. The distance to the other nucleon is then (s² + d²)½. Let that distance be denoted as z. The cosine of the angle between the radials and the direction of the force to the other nucleon of the opposite type is given by s/z. The radial component of that force is then

[Hq*exp(z/s0)/z²]*(s/z) = Hq*s*exp(z/s0)/z³

The total radial force on a nucleon is then

F = Hq*exp(−s/s0)/s² + Hq*s*exp(z/s0)/z³
which can be represented as
F = Hq*exp(−s/s0)/s²[1 + (s/z)³]

The ratio (s/z) can be represented as

(1 + (d/s)²)

When the previous analysis is repeated with the additional factor of [1 + (s/z)³] the result is

σ*exp(−σ)[1 + (σ/ζ)³] = n²h²/(2mHqs0)

where ζ=z/s0. This is the quantization condition for σ. A comparison of the LHS of the above with the function σ*exp(−σ) is shown below. There are differences in value but the shapes are quite similar. However one significant aspect is that the first function suggests that σ would not go beyond 1.0. The best estimate of s0 is from the Yukawa relation and is 1.522 fermi. The best estimate of the separation distances of the nucleons in an alpha particle is 1.68 fermi, which makes σ equal to 1.1. The second function allows for values of σ which are greater than unity.

(To be continued.)

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